Question

The graphs $y = 3(x-h)^2 + j$ and $y = 2(x-h)^2 + k$ have $y$ -intercepts of $2013$ and $2014,$ respectively, and each graph has two positive integer $x$ -intercepts. Find $h.$

Answer

Solution:

Setting $x = 0$ gives $3h^2 + j = 2013$ and $2h^2 + k = 2014.$ Expanding, the first graph is $y = 3x^2 - 6hx + 2013,$ whose roots are positive integers with sum $2h$ and product $\frac{2013}{3} = 671 = 11 \cdot 61.$ Similarly the second is $y = 2x^2 - 4hx + 2014,$ with integer roots of sum $2h$ and product $\frac{2014}{2} = 1007 = 19 \cdot 53.$

The first pair of roots is $\{11, 61\}$ or $\{1, 671\},$ so $2h = 72$ or $672;$ the second pair is $\{19, 53\}$ or $\{1, 1007\},$ so $2h = 72$ or $1008.$ The only common value is $2h = 72,$ so $h = 36,$ which indeed gives $x$ -intercepts $11, 61$ and $19, 53.$

2014 AIME I Problem 6

Solution:

Problem 6 in Other Years